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Offline mreff555

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aromaticity of compounds and huckles rule
« on: February 19, 2009, 11:28:10 AM »
I'm studying aromatic compounds and im having trouble reasoning an answer to the a quiz I just took. the quiz listed four compounds and asked me to name the aromatic one. The quiz said that the correct answer is a cyclooctatetradiene dianion, formed by the loss of two protons. Am I completely retarded. An 8 carbon ring with conjugated double bonds. Anti aromatic. So how is it that deprotonating it twice changes this? Wouldn't that be 12 pi electrons? how would the arrangement work. There would be two Sp3 carbons? Can someone please explain this?

Offline azmanam

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Re: aromaticity of compounds and huckles rule
« Reply #1 on: February 19, 2009, 11:36:55 AM »
Quote
cyclooctatetradiene

tetradiene?

what were the other three choices?
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Offline mreff555

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Re: aromaticity of compounds and huckles rule
« Reply #2 on: February 19, 2009, 11:41:30 AM »

Sorry, mistyped it.  cyclooctatetraene C8H8 with two abstracted protons
Quote
cyclooctatetradiene

tetradiene?

what were the other three choices?

Offline azmanam

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Re: aromaticity of compounds and huckles rule
« Reply #3 on: February 19, 2009, 11:55:14 AM »
C8H8 --> minus 2 protons --> C8H6???  Really?

Not aromatic in my book for two reasons.  One, how are you going to remove one proton, let alone two.  Second, electrons need to be in a plane for the ring to be aromatic.  With 8 sp2 centers on cyclooctatetraene, the two unlucky carbon atoms with the full negative charge are going to put the electrons in the sp2 orbital formerly occupied by the proton.  that orbital is perpendicular to the ring.  carbon atoms have been known to re-hybridize to accommodate aromaticity, but I don't see it going to sp in this case.

What were the other 3 choices, it's possible the answer sheet is wrong.
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Offline mreff555

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Re: aromaticity of compounds and huckles rule
« Reply #4 on: February 19, 2009, 12:04:57 PM »
Question 5 -  Which one of the following is aromatic?
     
     1. the cyclopentadienyl cation, formed by loss of a hydride anion
     2. the cyclooctatetraene dianion, formed by loss of two protons
     3. cyclobutadiene
     4. the cycloheptatrienyl anion, formed by loss of one proton
     5. cyclopropene

C8H8 --> minus 2 protons --> C8H6???  Really?

Not aromatic in my book for two reasons.  One, how are you going to remove one proton, let alone two.  Second, electrons need to be in a plane for the ring to be aromatic.  With 8 sp2 centers on cyclooctatetraene, the two unlucky carbon atoms with the full negative charge are going to put the electrons in the sp2 orbital formerly occupied by the proton.  that orbital is perpendicular to the ring.  carbon atoms have been known to re-hybridize to accommodate aromaticity, but I don't see it going to sp in this case.

What were the other 3 choices, it's possible the answer sheet is wrong.

Offline aldoxime_amine

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Re: aromaticity of compounds and huckles rule
« Reply #5 on: February 19, 2009, 12:24:57 PM »
Wouldn't the cyclooctatetraene dianion be non-aromatic due to conformation of the molecule into a non-planar shape? :-\

Offline azmanam

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Re: aromaticity of compounds and huckles rule
« Reply #6 on: February 19, 2009, 12:35:20 PM »
Quote
due to conformation of the molecule into a non-planar shape

yes.

unless I'm looking at it wrong, none of those are aromatic.  C8H8 isn't even planar, and C4H4 is antiaromatic.  cyclopentadenyl ANION is aromatic, as is cyclopropenyl CATION, and cycloheptatrieneyl CATION.
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Offline tarkolaszlo

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Re: aromaticity of compounds and huckles rule
« Reply #7 on: September 28, 2010, 02:55:59 AM »
Huckel model of aromaticity is obsolete.
See ARKIVOC, 2008, XI, p. 24.

We can send freeware, by E-mail, DESCRIPT software, for various aromaticity computations for any thinkable chemical structures.

Tarko L. - Center of Organic Chemistry - Bucharest - Romanian Academy ltarko@cco.ro

Offline science123

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Re: aromaticity of compounds and huckles rule
« Reply #8 on: September 28, 2010, 08:55:26 AM »
C8H8 --> minus 2 protons --> C8H6???  Really?

Not aromatic in my book for two reasons.  One, how are you going to remove one proton, let alone two.  Second, electrons need to be in a plane for the ring to be aromatic.  With 8 sp2 centers on cyclooctatetraene, the two unlucky carbon atoms with the full negative charge are going to put the electrons in the sp2 orbital formerly occupied by the proton.  that orbital is perpendicular to the ring.  carbon atoms have been known to re-hybridize to accommodate aromaticity, but I don't see it going to sp in this case.

What were the other 3 choices, it's possible the answer sheet is wrong.

The book we used for Organic chem (Wade 6th ed.) says that "The cyclooctatetraene dianion has a planar regular octagonal structure with C-C bond lengths of 1.4 Angstroms, close to the 1.397 A bond lengths of benzene. Cyclooctatetraene itself has eight pi electrons, so the dianion has ten...The dianion is easily prepared because it's aromatic."

It shows that cyclooctatetraene was reacted with K to give the dianion.

Offline azmanam

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Re: aromaticity of compounds and huckles rule
« Reply #9 on: September 28, 2010, 09:51:40 AM »
Thanks for the reference.  After searching out the original article, it appears you are correct.  Cyclooctatetraene dianion is aromatic.  I was thinking about it wrong above.  I assumed two protons were removed in an acid/base reaction to give the dianion C8H62-.

Instead, when reacted with elemental potassium, two electrons are pumped in to the cyclooctatetraene conjugated system to give the 10 electron C8H82-.

This 10 electron, planar dianion is in fact aromatic.  I hadn't heard of this before, and I apologize for any confusion.

http://en.wikipedia.org/wiki/Cyclooctatetraene
http://dx.doi.org/10.1021/ja01499a077

(money quote from JACS paper: "These crystals are difficult to isolate, for on drying and exposure to air they explode. Solutions however. are stable.")
Knowing why you got a question wrong is better than knowing that you got a question right.

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